Early Universe 5. Cosmological redshift and the dynamics of a homogeneous expanding universe, part 1

On the website of free lectures MIT OpenCourseWare posted a course of lectures on cosmology by Alan Gus, one of the creators of the inflationary model of the universe.

Your attention is invited to the translation of the fifth lecture: "Cosmological redshift and the dynamics of a homogeneous expanding universe, part 1".


Today we finish the consideration of the kinematics of a uniformly expanding universe, which we discussed last time. The only question from this topic that we have not yet touched is the cosmological redshift. Then we move on to the dynamics of homogeneous expansion — how gravity affects the expansion of the universe. This will be the main topic of today's and next few lectures.

Cosmological time

Let me remind you that at the end of the last lecture it was about synchronizing clocks in the coordinate system, which we will use to describe a uniformly expanding universe. Recall that we have introduced related coordinates that expand with the universe. We will assume that the universe is perfectly uniform and isotropic, and all objects rest in this coordinate system.

In the real universe there is some movement of matter relative to this coordinate system, because the universe is not completely homogeneous. But now we will work with an approximation in which our model universe is absolutely homogeneous and all matter rests relative to the expanding coordinate system.

Now let us recall how we determined the cosmological time at the last lecture. Imagine that at every point in the universe there are clocks that are at rest with respect to matter, and hence the expanding associated coordinate system. All such clocks measure local time, and we want to agree on their synchronization. Last time, we found out that we can synchronize clocks, if there are any cosmological phenomena that can be seen from anywhere in the universe, and that change with time. We gave two examples: one is the change in the Hubble constant, which can be measured locally and agree to set its clock to zero when the Hubble constant takes a certain value.

The second example is the temperature of the cosmic microwave background radiation. In our model universe, you can agree to set your clock to zero when the temperature of the cosmic background radiation reaches 5 degrees or any given number. If there are similar phenomena, and they are in our universe, then you can synchronize all the clocks. It is important to understand that once the clock is synchronized once, it will remain synchronized due to our assumption of homogeneity. That is, if everyone agrees that the temperature of the cosmic background radiation at zero time is 10 degrees, and everyone will wait 15 minutes, everyone will see the same temperature drop during this period of time, otherwise it would be a violation of our absolute homogeneity hypothesis .

STUDENT: Is it true that the background radiation temperature is the same for all inertial observers?

TEACHER: It is not exactly the same for different inertial observers. It is the same for the privileged class of observers who are at rest relative to the average distribution of matter and, therefore, relative to the associated coordinate system. If you begin to move through the cosmic background radiation, you will no longer see a uniform temperature distribution. You will see the radiation more hot in one direction and colder in the opposite. In fact, as I mentioned, we see this effect in our real universe. We, apparently, are moving relative to the cosmic background radiation, from about 1/1000 of the speed of light. So the radiation temperature is not invariant with respect to the movement of the observer.

You can ask another question - is the temperature of the background radiation the same in different places of the visible universe? As far as we can judge, yes. There is a direct way to measure the temperature of the background radiation, which we may discuss later in the course, by observing certain spectral lines in distant galaxies. In some galaxies where these lines are visible, one can directly measure the temperature of the cosmic microwave background radiation. In our model, we assume complete homogeneity, that everything is the same everywhere. Although the homogeneity in the real universe is not complete, there is strong evidence of the approximate homogeneity of our universe.

STUDENT: If some observers live near black holes, will this affect clock synchronization for such observers?

TEACHER: Of course it will. You can cosmologically synchronize clocks, only by assuming that the universe is absolutely homogeneous. As soon as irregularities appear, such as black holes, or even just stars like the Sun, they create deflections that prevent the clocks from synchronizing with each other. As soon as mass concentration appears, homogeneity becomes only approximate. But these deviations are small. The deviations arising from the Sun are on the order of one million. Therefore, in a very good approximation, the universe is described by our homogeneous model. Although, if you get very close to one of the supermassive black holes located in the centers of galaxies, it turns out that it has a very strong influence on the course of your clock.

Cosmological redshift

The next topic, as I promised, is the cosmological redshift. In the third lecture, we talked about the Doppler shift for sound waves and the relativistic Doppler shift for light waves, taking into account the special theory of relativity. However, cosmology is not fully described by the special theory of relativity, although the special theory of relativity is used to describe local events in cosmology. The special theory of relativity does not include the effects of gravity, and on a global scale, the effects of gravity are very important for cosmology. Therefore, the special theory of relativity is not enough to understand many properties of the universe, including the cosmological redshift. However, it turns out that there is a way to describe the cosmological redshift, which explains it even easier than the special theory of relativity. I will first describe it, and then we will talk about how this very simple-looking result correlates with the conclusion of the special theory of relativity, which must also be correct, at least locally.

So, suppose we are looking at a distant galaxy, and the light is emitted from a source in this galaxy. We want to understand what is the relationship between the frequency of light during emission and the frequency that we will see when receiving light.


To imagine this situation, let's introduce a coordinate system, $$$x$ . This will be our companion coordinate system. $$$x$ measured in divisions. We will place ourselves at the origin, and a distant galaxy at some distance from us. She has a specific coordinate $$$l_c$ ( $$$c$ denotes concomitant). $$$l_c$ - this is a concomitant distance from us to the galaxy. Physical distance which we call $$$l_p$ ( $$$p$ means physical) depends on time because the universe is expanding. As we said $$$l_p (t) = a (t) l_c$ . Scale factor $$$a (t)$ which depends on time, is multiplied by the accompanying distance, which does not depend on time. Thus, the physical distances simply increase in proportion to the scale factor. $$$a (t)$ .

Suppose now that the galaxy emits a light wave, and we are trying to determine the distance between the crests of the wave, which is equal to the wavelength. Since we will be interested only in ridges, we will simply imagine that each ridge is an impulse, and what happens between them does not interest us. We will follow the successive light pulses emitted by the galaxy.

It is important that we know for our model how light waves propagate in the associated coordinate system. If a $$$x$ - the accompanying coordinate, then $$$dx / dt$ - the accompanying speed of light, which is equal to the normal speed of light $$$c$ but divided by the scale factor:

$$

$$\ frac {dx} {dt} = \ frac c {a (t)}$$

The scale factor here plays the role of converting meters into divisions. $$$c$ measured in meters per second. Sharing $$$c$ on $$$a (t)$ we get the speed of light in tapers per second, just like we wanted, because $$$x$ measured not in meters, but in divisions. A division is an arbitrary unit that we choose to describe our associated coordinate system.

An important feature of this equation is that the speed of light in the associated coordinate system depends on time, but does not depend on $$$x$ . Our universe is homogeneous, so all points $$$x$ are equivalent. Therefore, at each moment in time two light pulses will move with the same accompanying speed, no matter where they are. That's all we need. The first impulse leaves a distant galaxy and moves towards us, the second impulse follows the first. The second impulse at any moment of time will move with the same accompanying speed as the first impulse, even if its accompanying speed changes with time.

This means the following. The accompanying pulse rate can vary with time, but as long as they both move at the same accompanying speed, at any moment in time they will be exactly at the same distance from each other in the accompanying coordinate system. $$$Δx$ The accompanying distance between two pulses does not change with time. If the accompanying distance does not change with time, and the physical distance is always equal to the product of the scale factor and the accompanying distance, then the physical wavelength of the light pulse will simply stretch in proportion to the scale factor. The wavelength will increase with the expansion of the universe, just like any other distance in our model of the universe will increase with the expansion of the universe. This is a key idea, it is very simple, and it contains everything.

That $$$Δx$ constantly means $$$Δl$ , physical distance, proportional $$$a (t)$ and this means that the wavelength of light $$$λ$ , as a function of t, is proportional $$$a (t)$ .

The wavelength is related to the period of the wave by the ratio $$$λ = cΔt$ . The wavelength is the distance that a wave travels in one period. Therefore, if $$$λ$ proportional to $$$a (t)$ , then $$$Δt$ wave period will be proportional $$$a (t)$ . Therefore:

$$$$ display $$ \ frac {Δt_ {det.}} {Δt_ {ist.}} = \ frac {λ _ {det.}} {λ _ {ist}} = \ frac {a (t_ {det. )}} {a (t_ {source)}} $$ display $$

.

Thus, the ratio of wavelengths is simply the number of times the universe has stretched. It is equal to the ratio of scale factors in the initial and final time. We determined the redshift using the wave period. The ratio of the periods, or the ratio of wavelengths, or the ratio of scale factors, is 1 + z.

$$

$$1 + z = \ frac {a (t_ {reg.)}} {A (t_ {ist.}}}$$

Relationship of cosmological redshift with special theory of relativity
How is the cosmological redshift associated with the redshift of the special theory of relativity, the formula for which we derived earlier? Our result differs in two respects from the calculation we did in the third lecture. The first reason that is important to us is that cosmological calculation takes into account effects that were not taken into account by previous calculations. In particular, despite the fact that we got an answer using a very simple kinematic argument, in which, at first glance, there is practically no mathematics, it is actually very strong, because it takes into account not only the special theory of relativity, but also the general theory relativity. It includes all the effects of gravity. Gravity does not affect the fact that the associated speed of light is $$$c / a (t)$ . This is just a unit transformation, together with the fundamental physical assumption that the speed of light is always equal to $$$c$ regarding any observer.

Therefore, when we consider gravity, this relationship continues to be preserved, and this was the only thing we used, so that gravity cannot affect the answer. Did we miss something from the special theory of relativity? I did not take into account the time dilation, which was crucial for our relativistic redshift calculation.

Did I make a mistake? Do I need to add time dilation somewhere? Not really. We had two clocks involved in our calculation: clocks in the galaxy and our clocks, which we used to measure the radiation period and reception period. But both of these clocks are at rest with respect to a local substance, even though they are moving relative to each other. Therefore, by definition, they measure cosmological time. Cosmological time is a very peculiar kind of time, it is not time in any inertial system. Clocks move relative to each other, therefore, if we determined the time in the inertial system, such clocks could never be synchronized and the time on them would not coincide.

But in the system of cosmological time, they show the same time. Since every watch is at rest relative to a local substance, they measure $$$t$ cosmological time. Thus, no time delay is needed. The point is not that we have forgotten him, he is not there. It is not used in calculations.

Thus, the result obtained, no matter how simple it may seem, actually fully covers the effects of both the special theory of relativity and gravity. Let me say that it’s not obvious where gravity is here. I told you that the result includes all the effects of gravity. Where is gravity hidden? I want to ask this question to you. How does gravity affect calculations, despite the fact that I did not mention gravity when I did the calculation?

STUDENT: Through the scale factor.

TEACHER: Right, through the scale factor. We have not talked about how to change $$$a (t)$ . Change $$$a (t)$ will obviously include the effects of gravity. That is why our result depends on gravity, although we did not need to use or mention gravity to get an answer. The answer for the cosmological redshift is so simple because $$$a (t)$ already includes a lot of information. We just used this to get a very simple expression depending on $$$a (t)$ , without saying anything about how we will calculate $$$a (t)$ . This is the first difference.

Another important difference between these two calculations is the variables used in the response. There may be several different answers to the same question, depending on which variables are used. In this case, we express the redshift. $$$z$ for objects resting in the associated coordinate system. Calculation in the special theory of relativity, on the other hand, gives z depending on the velocity measured in the inertial coordinate system. Thus, the results are expressed in completely different terms.

What happens if we try to compare the answers we received for relativistic and cosmological redshift? There is only one case in which it is legitimate to compare them. Since the calculation we have just done includes the effects of gravity, and the calculation using the special theory of relativity does not include the effects of gravity, the only case in which we can compare them and make sure that they match is the case when gravity is negligible.

One can consider a cosmological model where gravity is small; there is no contradiction in this. If gravity is negligible, how will it behave $$$a (t)$ ? If there is no gravity, $$$a (t)$ should be linearly dependent on $$$t$ . This means that all speeds are constant. Thus, in the particular case of the absence of gravity, $$$a (t)$ grows linearly with time. In this case, you can always make the constant that is added to the linear term become zero simply by setting the zero time at the moment when $$$a (t)$ equals zero. Thus, in the absence of gravity, it can be said that $$$a (t)$ should be proportional to t.

Then for this particular case, our two calculations must match. You can check it yourself. This is not so easy, it will require an understanding of the relationship between the two coordinate systems. The answer to the special theory of relativity is given in an inertial coordinate system, which in the presence of gravity does not exist at all. It is associated with the expanding coordinate system in a complex way, due to the slowing down of time and the Lorentz contraction associated with the movement that occurs in the expanding universe.

You will need to figure out the relationship between these two coordinate systems. When you do this, and compare the answers, you will find that they are exactly the same. All this agrees perfectly with the special theory of relativity, in the particular case where gravity is absent.

Newton and the static universe

We discussed everything that I wanted to talk about the kinematics of a uniformly expanding universe, now we are ready to move on to the dynamics. We need to figure out how gravity affects the universe in order to be able to calculate how $$$a (t)$ changes over time. It will be the only goal to understand the behavior. $$$a (t)$ .

This question, in a sense, goes back to Isaac Newton. I want to note that one of the interesting things in cosmology is that, if you look at the history of cosmology, many great physicists made big mistakes when trying to analyze cosmological issues. Today we will discuss one of Newton's mistakes. Even great physicists like Newton could make stupid mistakes. He really made a stupid mistake analyzing the cosmological consequences of his own theory of gravity.

Newton, like everything before Hubble, believed that the universe was static.He represented the universe as a static distribution of stars scattered in space. At the beginning of his career, as far as I know history, he assumed that the distribution of stars was finite in infinite space. But at some point he realized that if there is a finite distribution of mass in empty space, and all matter is attracted to each other with a force of attraction inversely proportional to the square of the distance, which he knew as he opened it, as a result everything should shrink the point. He decided that his assumption was not working, but still he was sure that the universe was static, because everything looked static, the stars did not move anywhere.

So he wondered what could be changed, and decided that instead of assuming that the stars form a finite distribution, it is better to assume that they are distributed infinitely in space. He reasoned as follows, and this is precisely the fallacy that if the stars fill infinite space, then even if they all attract each other, they will not have a preferred direction for movement. Since they will not have a preferred direction for movement, because they are affected by attraction from all sides, they remain in place. Thus, he believed that an infinite, uniform distribution of matter would be stable, that there would be no gravitational forces arising in such an infinite distribution of mass.

He apparently heard different arguments in favor of this. One of the arguments that the infinite distribution would be stable was the argument that an infinite force pulling it to the right acts on a particle and an infinite force pulling it to the left. Since both are endless, they neutralize each other. Newton did not accept this argument. He was sophisticated enough to understand that infinity minus infinity is not necessarily zero. However, Newton was convinced that the infinite distribution of mass would be stable.

The argument that convinced him was not the infinity of matter on each side, but symmetry. The argument he took was that if you look at any point of this infinite distribution, if you look around this point, all directions will look the same, with matter extending to infinity, and therefore there will be no direction in which the force should act on any particular particle. And if a force has no direction, it should be zero. This was the argument Newton was taking.

We will now discuss this in more detail and try to understand how modern scientists relate to this argument. By the way, I want to mention the historical fact. Newton's argument, as far as I know, has not been doubted by anyone for hundreds of years, right up to Albert Einstein. Albert Einstein, trying to describe cosmology with the help of his new general theory of relativity, was the first person to realize that even if you have an infinite distribution of mass, it collapses. Einstein realized that the same thing would happen in Newtonian physics, this is not a feature of the general theory of relativity. It just historically took the creation of a general theory of relativity to make people rethink and understand that Newton was wrong.

The impossibility of the existence of a static universe

The difficulty of trying to analyze a problem in the way Newton did is that Newton considered gravity as a force acting at a distance. If we have two objects located at a distance$$$r$ from each other, they will attract each other with a proportional force $$$1/r^2$ .Since the time of Newton, other ways of describing Newtonian gravity have been invented, which make things much more clear. The difficulty in using Newton's description is that if we try to add up all these forces, proportional$$$1/r^2$, we get divergent amounts, the interpretation of which we need to understand. But in order to understand that Newton was wrong, the easiest way is to look at Newton's other formulations of gravity. I will describe two of them, with both of which you may already be familiar with.

I will describe the first by analogy with Coulomb's law. Coulomb's law is really the same as the law of gravity. Coulomb's law says that any charged particle creates an electric field that is equal to the charge divided by the distance in the square and multiplied by the unit vector directed from the charge.

$$

$$\vec E=\frac q{r^2} \hat r$$

This is Coulomb's law. Sometimes there is a constant in it, depending on the units in which it is measured.$$$q$but it doesn't matter to us. I assume that we use this equation, where the constant is equal to 1.

As is known, the Gauss law can be obtained from the Coulomb's law . If the Coulomb's law is true, then we can definitely say what is the integral of the electric field flux over any closed surface. It is proportional to the total amount of charge inside the surface.

$$

$$\oint\limits_S \vec E \cdot d\vec a = 4πq_{}$$

Where $$$q_{}$equal to the total amount of charge inside a closed surface.

You can write down the law of Newton, almost the way Newton formulated it. You can express the gravitational acceleration at a given distance from the object:

$$

$$\vec g=\frac {-GM}{r^2} \hat r$$

This is the same inverse square law and is similar to Coulomb's law, with the exception of the constant at the beginning. The constant has the opposite sign, which is important in some cases, but not now. It is important that this equation can be reformulated as the Gauss law, and it is called the Gauss law of gravity. The only thing that makes it different is the constant in front:

$$

$$\oint\limits_S \vec g \cdot d\vec a = -4πGM_{}$$

Now let's take the uniform distribution of matter that Newton considered. Newton argued that you can take a uniform distribution of matter, filling all the infinite space, and it will be static, that is, there will be no acceleration. The lack of acceleration in Newton’s language means that$$$\vec g$must be everywhere zero. But from the last formula it follows that if$$$\vec g$ everywhere is zero, the integral of $$$\vec g$over any closed surface will also be zero, and, therefore, the total mass enclosed within this surface must also be zero. But if we have a uniform mass distribution, then the total enclosed mass, of course, will not be equal to zero for any non-zero volume. Thus, it is clear that the statement that the system is static, directly contradicts the Gauss formulation of the law of Newton.

Just for the sake of interest, I will give you another similar argument, using another more modern formulation of Newtonian gravity. If you have not met her and do not understand what I am talking about, do not worry, it is not so important. For those of you who know her, I will bring her. Another way of formulating Newton's gravity is to introduce a gravitational potential. I will use the letter$$$φ$for gravitational potential. It is associated with gravitational acceleration as follows:$$→g=−→∇φ$\vec g = - \vec ∇φ$ where $$→∇φ$\vec ∇φ$Is a gradient $$$φ$ . Gradient $$$φ$ equal to the unit vector $$$\hat i$ in the x direction multiplied by the derivative of $$$φ$ by $$$x$ plus a single vector $$$\hat j$ in the direction of the y-axis multiplied by $$$φ$ by $$$y$ plus a single vector $$$\hat k$ multiplied by the derivative of $$$φ$ by $$$z$ :

$$

$$\vec ∇φ = \hat i \frac {\partial φ}{\partial x} + \hat j \frac {\partial φ}{\partial y} + \hat k \frac {\partial φ}{\partial z}$$

Once we have determined the gravitational potential, we can write the differential form of the Gauss law, which becomes the so-called Poisson equation. It states that

$$

$$∇^2φ = 4πGρ$$

where ρ is the mass density, and $$$∇^2φ$ defined as the second derivative of $$$φ$ by $$$x$ plus the second derivative of $$$φ$ by $$$y$ plus the second derivative of $$$φ$ by $$$z$ :

$$

$$∇^2φ = \frac {\partial^2 φ}{\partial x^2} + \frac {\partial^2 φ}{\partial y^2} + \frac {\partial^2 φ}{\partial z^2}$$

This is called the Poisson equation. If the mass density is given, then the gravitational potential can be found, then its gradient can be calculated, and$$$\vec g$ .This is equivalent to other formulations of gravity. But this gives us another test of Newton's assertion that there is a uniform distribution of matter without any gravitational forces. If there is no gravitational force, then$$$\vec g$should be zero, as we said a minute ago. And from the fact that$$$\vec g$ is zero, it follows that the gradient $$$φ$equals zero.

If you look at the gradient formula, then this is a vector. For a zero vector, each of its three components must be zero, and, therefore, the derivative of$$$φ$ by $$$x$ will disappear, derived from $$$φ$ by $$$y$ derivative of $$$φ$ by $$$z$ will disappear. It means that $$$φ$must be constant everywhere, it has no derivative with respect to any spatial coordinate. Therefore, if$$$\vec g$ equal to zero then gradient $$$φ$ is zero and $$$φ$ is a constant in all space. If a $$$φ$ everywhere is constant, which takes place in the absence of gravity, then it is immediately obvious that $$$∇^2φ$must be zero, which means ρ must be zero, that is, there will be no mass density. But Newton wanted to have a non-zero mass density, a substance uniformly distributed over infinite space. This is another demonstration that Newton's argument was wrong.

Conditionally convergent integrals

So, we came to the conclusion that Newton was wrong, but we need to more thoroughly analyze Newton's argument in order to understand exactly where he was wrong. The next thing I want to discuss is the ambiguity associated with the addition of Newton's gravitational forces for an infinite universe. I mentioned that the real problem with Newton's calculations is that the amount he calculated is divergent, and you need to be careful when trying to calculate it.

To clarify this, I want to start with an example of an integral that gives an ambiguous value. I will introduce a couple of mathematical concepts. Let's imagine that we have some arbitrary function$$$f (x)$ where $$$x$will be just one variable.

We will generalize it into three dimensions, which we are interested in, but we will begin with one variable. If we have a function$$$f (x)$ we can consider the integral from minus infinity to infinity from $$$f (x)$ I will call him $$$I_1$ :

$$

$$I_1 = \int\limits_{-\infty}^\infty f(x)dx$$

It is this integral that is obtained by adding all the gravitational forces acting on the body. Now I want to consider the case when$$$I_1$is finite

I need to first define more precisely what I mean by$$$I_1$, integral from minus to plus infinity. We can define the integral from minus infinity to infinity as the limit of the integral from$$$-L$ before $$$L$ from $$$f (x)$ in which $$$L$ tends to infinity:

$$

$$I_1=\lim_{L \to \infty} \int \limits_{-L}^Lf(x)dx$$

We need to calculate the integral of $$$-L$ before $$$L$ . If we assume that $$$f()$is finite, the integral is also always finite. I will assume that the function itself$$$f (x)$ finite, we will only worry about the convergence of the integral with $$$L$tending to infinity. So for any given$$$L$the integral is a number. Then one may wonder whether this number tends to the limit, when$$$L$tends to infinity? If so, we will call it the value.$$$I_1$ .It is simply a definition of what we mean by an integral from minus infinity to infinity.

Now consider the case when this value exists when$$$I_1$less than infinity, that is, it has some finite value. But I also want to consider the integral, which I will call$$$I_2$ which is also defined as an integral from minus infinity to infinity but from the absolute value $$$f (x)$ :

$$

$$I_2 = \int\limits_{-\infty}^\infty |f(x)|dx$$

Now a bit of terminology. If a $$$I_2$ less than infinity, if it converges, then $$$I_1$called absolutely convergent . Absolutely convergent means that the integral converges even if the absolute value of the function is used. On the contrary, if$$$I_2$ diverges, but at the same time $$$I_1$ converges then $$$I_1$called conditionally convergent . Thus, if the integral of the function converges, but the integral of the absolute value of the same function does not converge, then this case is called conditional convergence.

The reason for this separation is that conditionally convergent integrals are very dangerous. They are dangerous because they are not well defined. You can get any value we want by adding the integrand in a different order. As long as we adhere to a particular order, which is implied in the symbol of the integral, we get a unique answer. But, if, for example, we just move the beginning of the integration, we can get another answer. Usually we do not expect this. Usually we simply integrate along the numerical line, no matter where we begin to calculate the integral. Thus, the result becomes much less definite when we work with conditionally convergent integrals.

Before turning to a specific integral that interests us, with which we try to add up the gravitational forces of the infinite distribution of matter, I will give an example of a very simple function that illustrates this ambiguity when the integral converges, but does not absolutely converge. You can get any answer we want by folding the parts of the integral in a different order. An example that I will consider is a function f (x), which is +1 if x> 0 and -1 if x <0. I do not indicate what it is equal to, if x = 0, when integrating it does not matter. The only point does not matter. You can take any value for the function when x = 0, it will not change anything.

If we integrate it symmetrically, following the definition of what we mean by integration from minus infinity to infinity, we get a complete reduction.

When we integrate from $$$-L$ before $$$L$, we get zero, because there is a complete reduction between the negative parts and the positive parts of the integral. Then, if we take the limit, when$$$L$tends to infinity, the zero limit will be zero. There is no ambiguity in this statement.

Thus, adding the parts of the integral in the indicated order we get the integral, which is equal to zero. But the result depends on the order in which we fold these pieces. In particular, if you simply change the beginning of integration, starting to move aside from the new beginning, we will get another answer. Let's look at the limit again when$$$L$ tends to infinity but instead of integrating from $$$-L$ before $$$L$ we will integrate from $$$a-L$ before $$$a+L$ .

This is actually the same integral, we just shifted to the right our beginning of integration. In the particular case$$$a$ equals zero, and we get the same as before, but if $$$a$ not zero, this means that our integral is calculated starting from $$$x=a$ and not from $$$x = 0$ .

First we must calculate the integral of $$$a-L$ before $$$a+L$ and then take the limit when $$$L$strive towards infinity, and see what we get.

Easy to understand what we get. Once$$$L$ getting bigger $$$a$ , the answer no longer changes with increasing $$$L$ . When we increase $$$L$, we add some negative part to the left, and the same positive to the right, and they neutralize each other. When$$$L=a$ the integral will be from 0 to 2 $$. In the integral there will be only positive values ​​of the function, the integration interval will be equal to 2 $$This means that the integral will be equal to 2 $$. For any large values $$$L$ the integral will be the same because, with increasing $$$L$as I said, we simply reduce the addition of positive values ​​on the right and negative values ​​on the left. Therefore, the limit for this integration has a well-defined value, which is 2$$ .

$$$a$- this is the number with which we started the integration, so it can be anything. We can choose$$$a$what we want. Thus, we can get any answer we want if we can add parts of the integral in any order. This is the fundamental uncertainty of conditionally convergent integrals. We will see that an attempt to add up the forces acting on a particle in an infinite distribution of mass is just such a conditionally convergent integral. Therefore, we can get any answer we want, and it will not mean anything unless it is done very carefully.

The problem of the addition of gravitational forces

Now I want to calculate the force acting on any particle in the infinite distribution of matter and show that I can get different answers, depending on the order in which I add the gravitational forces. In each example, I will add forces in a certain order, and I will receive a definite answer, but I will receive different answers, depending on which order of addition I choose.

Let's try to calculate the gravitational force at some point. $$$P$in the infinite distribution of matter. Substance fills the slide, and all space, to infinity. We will add the gravitational contribution of all this substance in a certain order.

In our first calculation, we add the gravitational forces from a substance located in concentric shells with a center at$$$P$ .First we take the innermost shell, then the second shell, the third shell, and so on. moving farther from the center. In this case, it is easy to understand that the force acting on a point$$$P$ calculated in this order is equal to 0, because for each shell $$$P$is in the center, and for reasons of symmetry, the forces must be compensated. In fact, it is known, and we will soon take advantage of this fact that the gravitational field of the shell inside the shell is zero. This was proved by Newton. And outside the shell, the gravitational field looks exactly the same as if all the substance of the shell was concentrated in its center. It is clear that in this case the gravitational force at the point$$$P$ equals 0.

Now we consider a more complicated case in which we also calculate the gravitational force at the point $$$P$ . But we will use spherical shells centered around another point, $$$Q$ . Now $$$Q$defines the shells we will use for folding forces. We also add forces from all shells from zero to infinity, i.e. add up all the forces at the point$$$P$from the whole infinite distribution of matter. But we will add these forces in a different order, because we will take in order the shells with the center in$$$Q$ . First we consider the contribution of the shaded area, which is all the shells around $$$Q$ having radii less than the distance from $$$Q$ before $$$P$ . For all these shells, the point $$$P$lies outside the shell. Therefore, each shell acts in the same way as a point mass equal to the total mass of the shell, concentrated at the point$$$Q$, the center of all shells. Thus, a substance that is in a shaded area contributes to the power at the point$$$P$ equal to the force that would create a shadowed mass if it were all concentrated at $$$Q$ .

On the other hand, all other shells will be shells for which $$$P$ is inside. $$$P$is no longer at the center of these shells, but Newton found out that it doesn’t matter. Inside the spherical shell, the gravitational force is zero anywhere, no matter how close it is to the boundary. All forces from different parts of the shell are exactly compensated. If you get closer to the border, then we can assume that there will be an attraction towards this border. Indeed, in this case, the force of attraction to a particular particle on this boundary becomes stronger, because it is proportional to$$$1/r^2$ .But as we approach the border, more and more matter is on the opposite side. And these two effects fully compensate each other. By the way, the fact that the force acting on a particle inside the shell is zero is easy to prove, using Gauss's law for gravity.

Therefore, the outer shell does not make any contribution. We found out that force is at$$$P$equal to the force created by the shaded mass. Gravitational acceleration at a point$$$P$ is determined by a simple formula: it is equal to G times the total mass of the shaded area divided by $$$b^2$ where $$$b$ equal to the distance between $$$Q$ and $$$P$ multiplied by a unit vector directed from $$$Q$ to the side $$$P$ :

$$

$$\vec g = \frac {GM}{b^2}\hat e_{QP}$$

And this is a non-zero value. Thus, we get a zero or non-zero result, depending on the order in which we add forces from our infinite distribution of matter. In addition, we can get any answer, because we can choose$$$b$whatever you like. The answer depends on$$$b$ and becomes arbitrarily large as you increase $$$b$ . It may seem that the answer decreases with increasing $$$b$ but in fact it grows because the mass $$$M$ grows like $$$b^3$ . We can gain strength in any direction by choosing $$$Q$ in the right direction from $$$P$ .We can really get any answer using this way of combining forces.

The problem is that these shells do not really exist. We just mentally work with these shells. The substance is evenly distributed and there are no shells. Shells are purely mental objects that should not affect the answer. They only determine the order in which we sum the gravitational forces.